Question

Suppose the ends of the coil in the previous problem are connected to a resistance of 100 Ω. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.

Answer 1

Given:
T = 1 minute
Heat produced in the circuit is calculated using the following relation:
$H=\underset{0}{\overset{T}{\int }}{i}^{2}Rdt$
$\Rightarrow H=\underset{0}{\overset{1\min }{\int }}\frac{B^2{A}^2{\omega }^2}{R^2}\sin \left(\omega t\right)Rdt$
$$\begin{gathered} =\frac{B^2{A}^2{\omega }^2}{2R}.\underset{0}{\overset{1\min }{\int }}\left(1-\cos 2\mathrm{\omega }t\right)dt \\ =\frac{B^2{A}^2{\omega }^2}{2R}{\left(1-\frac{\sin 2\omega t}{2\omega }\right)}_0^{1\min } \\ =\frac{B^2{A}^2{\omega }^2}{2R}\left(60-\frac{\sin 2\times 80\times 2\mathrm{\pi }/60\times 60}{2\times 80\times 2\mathrm{\pi }/60}\right) \\ =\frac{60}{2R}\times {\pi }^2{r}^4\times B^2\times {\left(80\times \frac{2\pi }{60}\right)}^2 \\ =\frac{60}{200}\times 10\times \frac{64}{9}\times 10\times 625\times {10}^{-8}\times {10}^{-4} \\ =\frac{625\times 6\times 64}{9\times 2}\times {10}^{-11}=1.33\times {10}^{-7}\mathrm{J} \end{gathered}$$